# Covariant derivative wrt a superscript sign problem

1. Aug 9, 2014

### cr7einstein

Dear all,
I was reading this https://sites.google.com/site/gener...-the-covariant-derivative-of-the-ricci-tensor, and it said that if you take the covariant derivative of a tensor with respect to a superscript, then the partial derivative term has a MINUS sign. How? The Christoffel symbol should have a minus sign, but I don't understand how does the partial derivative get one?

Also, does covariant derivative always have an index opposite to that of the tensor(e.g. a contravariant tensor will be differentiated wrt a covariant tensor, and a covariant tensor wrt to a covariant index)? If so, why? Is there a relation between the two(which the minus sign mentioned above indicates)?

2. Aug 9, 2014

### WannabeNewton

None of that is correct actually. $\nabla^{\mu}A^{\nu} = g^{\mu\delta}\nabla_{\delta}A^{\nu} = g^{\mu\delta}\partial_{\delta}A^{\nu} + g^{\mu\delta}\Gamma^{\nu}_{\delta \sigma}A^{\sigma}$ and similarly for tensors of arbitrary rank, so it depends entirely on $g_{\mu\nu}$.

No.

3. Aug 9, 2014

### dextercioby

I'm afraid you didn't choose a bright source for reading. So, in general relativity there's no $x_{\mu}$ (and in special relativity shouldn't be either). Next:

$$\nabla^{\mu} T_{\alpha} = g^{\mu\beta}\nabla_{\beta}T_{\alpha}$$

is just a shorthand whenever necessary. Because the nonmetricity is 0, then you can play around freely with the position of the index in the covariant derivative. The metric also allows you to play with the index position also for the tensorial objects being differentiated:

$$\nabla^{\mu}T_{\alpha}^{~~\gamma} = g_{\alpha\delta} g^{\mu\lambda}\nabla_{\lambda}T^{\delta\gamma}$$

Relativists call this 'idex gymnastics'.